# Derivation of Kepler's third law using Virial theorem

I am familiar with the long derivation of Kepler's third law using the equations of motion. One starts with

$$$$\dot{r}=\sqrt{\frac{2}{m}\big[E-V(r)\big]}$$$$

and integrates to get

$$$$\Delta t =\int_{r_0}^r \frac{1}{\sqrt{\frac{2}{m}\big[E-V(r)\big]}}dr$$$$

Then in the particular case of a full orbit, one gets

$$$$T=\sqrt{\frac{a^3}{GM}}2\pi$$$$

I'm wondering if there's a derivation of the relation between period $$T$$ and semi-major axis $$a$$ that doesn't involve complete solving the equations of motion but only relies on the Virial theorem.

I'm familiar that in the case of circular orbits it's very easy to get Kepler's third law without using equations of motion. I'm wondering if one can get the general Kepler's third law that works for ellipses as well.

• The equations of motion of a system of gravitating masses have scale invariance. Kepler's third law easily follows from this fact for geometrically similar ellipses.
– Gec
Oct 4, 2023 at 6:45

An elementary derivation I know is the one proposed by Ellis D. Noll, Phys. Teach. 34, 42–43 (1996) https://doi.org/10.1119/1.2344336 that I'll briefly summarize here.

Kepler's second law is derived by the conservation of angular momentum $$\bf L$$. Using $$\bf r$$ and $$\bf v$$ to indicate the position and velocity of the planet, $$M$$ and $$m$$ the masses of the two bodies, $$\mu = \frac{mM}{m+M}$$ the reduced mass, and $$A(t)$$ the area of the sector of ellipse limited by the position vector at a starting time $$t_0$$ and time $$t$$, we have $$dA = \frac12 |{\bf r} \times {\bf v}| dt = \frac{L}{2 \mu} dt.$$ Being $$L$$ constant, the integration over a period $$T$$ gives the area of the ellipse of semiaxes $$a$$ and $$b$$: $$\pi a b=\frac{L}{2 \mu} T.\tag{1}$$ At the minimum distance between the two bodies, the velocity is perpendicular to the position, and we can write $$L = \mu r_{min}v_{min}.$$ Squaring such relation and dividing by 2, we get a relation between minimum distance, angular momentum, and kinetic energy at the minimum distance ($$K_{m}$$): $$K_m = \frac{L^2}{2 \mu r^2_{min}}.$$ By evaluating the conserved energy at the positions of minimum and maximum distances, we have $$\frac{L^2}{2 \mu r^2_{min}}-\frac{GMm}{r_{min}}= \frac{L^2}{2 \mu r^2_{max}}-\frac{GMm}{r_{max}}$$ from which, using the fact that $$r_{min}+r_{max}=2a$$, we get $$L^2=\frac{GMm\mu}{a}(r_{min}r_{max}).$$ At this point we can use the relations \begin{align} r_{min}&=a(1-e)\\ r_{max}&=a(1+e)\\ b^2&=a^2(1-e^2) \end{align} that give $$r_{min}r_{max}=b^2$$. Then, we can write $$L^2=\frac{GMm\mu b^2}{a}.\tag{2}$$ Squaring equation ($$1$$) and using equation ($$2$$), we finally get Kepler's third law in its general form. In particular, for $$M>>m$$, $$\mu \simeq m$$ and we get $$T^2 = \frac{4 \pi^2}{GM}a^3.$$

In this answer I will first give a general discussion of the Virial theorem, and then I will show how to obtain Kepler's third law.

It may well be that what is given below is in fact what you are alluding to in stating "in the case of circular orbits it's very easy to get Kepler's third law without using equations of motion".

There are two aspects to the Virial Theorem. One aspect is statement about interconversion of kinetic energy and potential energy, the other aspect is about averaging.

An example of a discussion where that distinction is made is the one by John Baez. The virial theorem:

We have that the distance dependence of some force law can be any power of the distance value. $$r$$.

$$... , \ r^{-3}, \ r^{-2}, \ r^{-1}, \ \text{constant force}, \ r^1, \ r^2, \ ...$$

As we know: out of that entire range there are two force laws that give rise to orbits that loop back on themselves: Hooke's law, and when the force law is an inverse square law. I will narrow down to those two.

Also, for simplicity I narrow down to the case of a small object circumnavigating a far more massive object, such that in the calculation it is isn't necessary to use reduced mass.

For $$p$$ negative or positive: a distance dependent force law:

$$F = k r^p \tag{1}$$

In a bound system (such as a gravitationally bound system) the velocity of an object in circular orbit is such that the exerted force is the amount of centripetal force that is required. As we know, for a given velocity $$v$$ the required centripetal force is $$\frac{m v^2}{r}$$

$$\frac{m v^2}{r} = k r^p \tag{2}$$

A this point we take advantage of the fact that force-that-sustains-circumnavigating-motion and kinetic energy have the quantity $$m v^2$$ in common. We multiply both sides of (2) with $$\tfrac{1}{2}r$$; that changes the left hand side of (2) into the expression for kinetic energy.

$$\frac{1}{2}m v^2 = \frac{1}{2} k r^{p+1} \tag{3}$$

The multiplication with $$r$$ had the effect of changing $$r^p$$ to $$r^{p+1}$$; the exponent of $$r$$ was raised by one unit. That is the same as what happens in integration of the force over distance.

Example:

Hooke's law:

$$\frac{m v^2}{r} = k r^1 \tag{4}$$

$$\frac{1}{2}m v^2 = \frac{1}{2} k r^2 \tag{5}$$

Therefore in the case of Hooke's law we have:

$$E_k = E_p \tag{6}$$

What that says is: if you have a bound system and the centripetal force is according to Hooke's law, then when you increase the radius of circumnavigating motion the kinetic energy and the potential energy will remain in a 1-to-1 ratio.

Next the inverse square law of gravity:

$$\frac{m v^2}{r} = GM m \ r^{-2} \tag{7}$$

$$\frac{1}{2}m v^2 = \frac{1}{2} GM m \ r^{-1} \tag{8}$$

In order to treat the right hand side of (8) as an expression for potential energy we need to introduce a minus sign. That minus sign expresses that when you shift the altitude of an orbit the kinetic energy and potential energy are counter-changing.

$$2 E_k = -E_p \tag{9}$$

To go from (8) to Kepler's third law we start as follows: we drop the factor $$m$$, and we move a factor $$r$$ to the other side

$$v^2 r = GM \tag{10}$$

Then the substitution: $$v = \omega r$$ results in an expression that is equivalent to Kepler's third law:

$$\omega^2 r^3 = GM \tag{11}$$

I haven't looked at how to generalize (11) to also cover eccentric orbits (which is what you are asking about specifically).

What I expect:
Compare a circular orbit with an eccentric orbit of the same total energy. I expect that the period of the eccentric orbit will be close to the period of the circular orbit. That is: I expect that when the motion of an eccentric orbit is averaged out (11) will obtain.